Physics 115C Homework 3

Transcription

1 Physics 115C Homework 3 Problem 1 In this problem, it will be convenient to introduce the Einstein summation convention. Note that we can write S = i S i i where the sum is over i = x,y,z. In the Einstein summation convention, we leave out the overall summation sign, and assume the convention that if an index is repeated in an expression, there is an implied summation over it, like so: i S i i i S i This will make our expressions easier to work with (and will save some typing for me. Now, let s start. Recall that the commutation relations for angular momentum are [ i, j ] = iǫ ijk k [, i ] = where ǫ ijk is the evi-civita symbol, and identical expressions hold for S and J (and there is an implied sum over the index k!. Using these commutation relations along with the definition J = +S, let s compute the commutator of S with J,, S, J z, z, and S z : [ S, ] = [ S,S ] = These two are trivial, since and S commute with all of their components, and therefore also with S. Next, [ S,J ] = [ S,J J] = [ S,(+S (+S] = [ S, +S + S] = [ S, ]+[ S,S ]+[ S, S] = [ S, ]+[ S,S ] = 1

2 To get the the fourth line, I used the fact that since S and commute, S = S ; to get to the fifth line, I used the fact that S commutes with itself; and to get to the last line, I used the fact that commutes with all the components of S and. Continuing, [ S, z ] = [ i S i, z ] = i [S i, z ]+[ i, z ]S i = +iǫ izk k S i = iǫ zki k S i = i( S z To get to the third line, I used the fact that the components of S and commute with each other; to get to the fourth line, I used the antisymmetry of the evi-civita symbol under exchange of any two indices (e.g. ǫ ijk = ǫ jik = ǫ jki, etc.; and to get to the last line I used the expression for the dot product Moving on, Finally, (A B i = ǫ ijk A j B k [ S,S z ] = [ i S i,s z ] = i [S i,s z ]+[ i,s z ]S i = iǫ izk i S k + = iǫ zik i S k = i( S z [ S,J z ] = [ S, z +S z ] = [ S, z ]+[ S,S z ] = i( S z i( S z = where I made use of our results above. Therefore, J,,S, and J z commute with S, but not z and S z. Since the spin-orbit coupling perturbation is proportional to S, this is why the good basis states are those labeled by n,l,s,j, and m j, but not those labeled by n,l,s,m l, and m s.

3 Problem (Griffiths 6.1 From Griffiths equations 6.75 and 6.76, we have that the first-order corrections to the energy in the weak-field Zeeman effect are E (1 Z = µ Bg J B ext m j where µ B is the Bohr magneton, B ext is the external electric field (along which we align our z-axis, and g J is the andé g-factor: g J = 1+ j(j +1 l(l+1+3/4 j(j +1 (incidentally, Griffiths derivation of the andé g-factor gives a good physical idea of what s going on, but isn t very mathematically rigorous. A more rigorous derivation of equation 6.73 makes use of the Wigner-Eckart theorem. Now, the eight n = states of the hydrogen atom have s = 1/ and l = or l = 1. For l =, the only possible value of j is 1/; for l = 1, the possible values of j are j = 1/ and j = 3/. Thus there are two states with l = (corresponding to quantum numbers m j = ±1/ and six states with l = 1 (j = 1/ and m j = ±1/, and j = 3/ and m j = ±1/,±3/. et s label these states as follows: using the notation ljm j, we define the eight states as 1 =,1/,1/ =,1/, 1/ 3 = 1,1/,1/ 4 = 1,1/, 1/ 5 = 1,3/,3/ 6 = 1,3/,1/ 7 = 1,3/, 1/ 8 = 1,3/, 3/ Now, let s calculate the g-factors for these states. For 1 and, we have g J = 1+ 1/(1+1/ +3/4 (1/(1+1/ = For 3 and 4, we have g J = 1+ 1/(1+1/ 1(1+1+3/4 (1/(1+1/ = 3 For the last four states, we have g J = 1+ 3/(1+3/ 1(1+1+3/4 (3/(1+3/ = 4 3 3

5 Here s a depiction of how the energies vary with the perturbing magnetic field B ext (not to scale, of course: 5

6 Problem 3 If a system begins at time t = in an eigenstate i of the unperturbed Hamiltonian, then the probability of measuring the system to be in some other state f at a later time t (after the perturbation has been turned on is c f (t, where c f (t = δ fi i t e i(e( f E( i t / f H (t i dt where H (t is the perturbing Hamiltonian. In our particular case, we have H (t = βxe t/τ To give this perturbation a physical interpretation, we might imagine that this would be the perturbation the oscillator would experience if it were placed in an exponentially decaying electric field(which might arise, for instance, inside of a capacitor as it was being discharge in anrcircuit. Wearefurthergiventhattheoscillatorbeginsinthegroundstate,so i =. Since the (unperturbed energies of the harmonic oscillator are given by E n ( = (n+1/ω, we have c n (t = δ n i = δ n iβ t t [ (( exp i n+ 1 e inωt e t /τ n x dt ω 1 ω t ] n H (t dt Since we re interested only in the probability of measuring the system to be in the nth state at large time, we might as well take the limit t : c n ( = δ n iβ n x e (1/τ inωt dt = δ n iβ n x e (1/τ inωt dt = δ n iβ [ n x 1 1/τ inω e (1/τ inωt = δ n iβ (1/τ inω n x To evaluate the matrix element n x, let s use the ladder operators: recall that we can express x = mω (a +a ] 6

7 Thus and therefore n x = = = c n ( = δ n mω n (a +a mω n 1 mω δ n1 mω iβ (1/τ inω δ n1 Note that this expression is zero unless n = 1 or n =, i.e. the probability of the system transitioning to any state higher than n = 1 is zero (at least to first order in perturbation theory. To get the probability of transition, we take the norm squared: the probability of no transition at all is P = c ( = 1 = 1 The probability of transitioning to the first excited state is P 1 = c 1 ( = iβ mω (1/τ iω = = β mω (1/τ +ω β mω(1/τ +ω Thus the probability of measuring the system to be in the nth state at large time is 1 n = β P n = n = 1 mω(1/τ +ω otherwise 7

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